Problem: A few families took a trip to an amusement park together. Tickets cost $$7.00$ each for adults and $$3.50$ each for kids, and the group paid $$45.50$ in total. There were $7$ fewer adults than kids in the group. Find the number of adults and kids on the trip.
Answer: Let $x$ equal the number of adults and $y$ equal the number of kids. The system of equations is then: ${7x+3.5y = 45.5}$ ${x = y-7}$ Solve for $x$ and $y$ using substitution. Since $x$ has already been solved for, substitute ${y-7}$ for $x$ in the first equation. ${7}{(y-7)}{+ 3.5y = 45.5}$ Simplify and solve for $y$ $ 7y-49 + 3.5y = 45.5 $ $ 10.5y-49 = 45.5 $ $ 10.5y = 94.5 $ $ y = \dfrac{94.5}{10.5} $ ${y = 9}$ Now that you know ${y = 9}$ , plug it back into ${x = y-7}$ to find $x$ ${x = }{(9)}{ - 7}$ ${x = 2}$ You can also plug ${y = 9}$ into ${7x+3.5y = 45.5}$ and get the same answer for $x$ ${7x + 3.5}{(9)}{= 45.5}$ ${x = 2}$ There were $2$ adults and $9$ kids.